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I was stuck on an assignment and I think I solved it, but I am not sure if I did it correctly.

This is the question:

Problem 3.8.5: Let $a$, $b$, $c$ be real numbers. Consider the equation $z = ax+by+c$. Prove that there are three 3-vectors $v_0$, $v_1$, $v_2$ such that the set of points $[x, y, z]$ satisfying the equation is exactly $\{v_0 + α_1 v_1 + α_2 v_2 : α_1 ∈ \mathbb R, α_2 ∈ \mathbb R\}$ (Hint: Specify the vectors using formulas involving $a$, $b$, $c$.)

What I did is:

1) I took the normal form, which is: $ax+by - z= -c$

2) I used some numbers, imagining they are on the plane, to prove it, like this:

The normal form: $[1, 2, -3]$ ($-3$ because $- z$ in the equation above)

$\mathbf x = [x, y, z]$

$\mathbf x_0 = [1, 2, 3]$

After that, I did:

$\mathbf x - \mathbf x_0 = [x-1, y-2, z-3]$

3) Now, time to use the dot product and multiply the normal form with the outcome of the last step. So: $\text{normal form} \cdot (x-x0)$

This gives me:

$$1 \cdot (x-1) + (2y - 2) + - 3 (z-3) = 0$$

Doing the algebra magic, I eventually get:

$$x + 2y - 3z = -4$$

Which is equivalent to $ax+by - z= -c$

My question:

is this correct and did I do it correctly?

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    First, although working through a specific example can be very helpful in unbdertanding what’s going on, it’s not a general proof. Second, what you’ve done is construct the equation of a specific plane from a normal to the plane and a point on the plane. The problem is asking you to do something else. You’re basically being asked to find a parameterization of the plane $z=ax+by+c$, which requires you to come up with *three* vectors. Hint: two of the vectors are parallel to the plane. – 2017-02-18
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    So... what you mean is that I didn't answer the question? If so, how should I answer it? I don't understand it... damn, this is hard for me. Can you help me with an example, please? – 2017-02-18
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    No, you didn’t answer the question. If you need an example, start from your last equation and try to find the three vectors that the question asks for. – 2017-02-19
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    You mean this one: ax+by−z=−cax+by−z=−c ? If so, should I just input some random numbers or so? The thing is that I just don't understand HOW to find the three vectors, because I don't know how to approach it. Can you give me tips, hints or just a rule to help me out? – 2017-02-19

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In this problem you’re being asked to find a parametrization $v_0+\alpha_1v_1+\alpha_2v_2$ of the general equation $z=ax+by+c\tag{*}$ of a plane in $\mathbb R^3$. There is in fact an infinite number of such parametrizations, but we’ll only produce one.

By setting $\alpha_1=\alpha_2=0$, we see that $v_0$ must be the position vector of a point on the plane. We can find such a vector by setting $x=y=0$, so that $z=c$, hence we can take $v_0=[0,0,c]$. Now, if $v$ is the position vector of any point on the plane, $v-v_0=\alpha_1v_1+\alpha_2v_2$, from which we can see that $v_1$ and $v_2$ must be parallel to the plane. This equation also tells us that we can think of our plane as the span of $v_1$ and $v_2$, which is a plane through the origin, translated by $v_0$. We can also see that in order to generate a plane, $v_1$ and $v_2$ must be linearly independent, i.e., not be scalar multiples of each other.

By rewriting equation (*) as $ax+by-z=-c$, we know that $[a,b,-1]$ is normal to the plane, so any vector that is orthogonal to this normal is parallel to the plane. It’s not hard to come up with $v_1=[1,0,a]$ and $v_2=[0,1,b]$ by inspection. (I’ll leave it to you to verify that they are indeed orthogonal to $[a,b,-1]$.) Putting this all together, we get for our parameterization $$[0,0,c]+\alpha_1[1,0,a]+\alpha_2[0,1,b]=[\alpha_1,\alpha_2,a\alpha_1+b\alpha_2+c].$$ One can see at a glance that this vector satisfies equation (*).